posted by Steve .
When reviewing the procedure the student found that 4.912 x 10-1 M barium nitrate solution had been used instead of that in the following procedure:
A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 5.912 x 10-1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004 x 10-1 M sodium Iodate, NaIO3
Calculate the theoretical yield of barium iodate monohydrate using 30.00 mL of 4.912 x 10-1 M barium nitrate solution
HERE IS THE QUESTION: SUPPLEMENTAL INFO IS ABOVE
Assume that, in the experiment described in Post-Laboratory Question 2, 125 mL of 25 C distilled water was used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (4 C).
The solubility of Barium iodate monohydrate in 25 C water is 0.028 g per 100 mL of water; in 4 C water, it is 0.010 g per 100 mL of water.
What mass of product would you expect to isolate?
I worked this a couple of days ago but that problem didn't included the water washing as described here. There is so much superfluous information here it makes the problem a little confusing. Here is what I will work with.
Calculate theoretical yield of Ba(IO3)2 using 30.00 mL of 0.4912M Ba(NO3)2 added to 50.00 mL of 0.9004M NaIO3 and we will assume this experiment was conducted in chilled (4C) water.
Ba(NO3)2 + 2NaIO3 + H2O ==> Ba(IO32)2.H2O + 2NaNO3
mols Ba(NO3)2 = M x L = estimated 0.015 but you should get a more accurate answer to this and all of the estimated answers that follow.
mols Ba(IO3)2 formed = estd 0.015 if this is the limiting reagent(LR).
mols NaIO3 = M x L = 0.045. mols Ba(IO3)4 formed from this would be 1/2 x 0.045 = estd 0.0225 so NaIO3 is in excess and about 0.015 mols Ba(IO3)2 will be formed.
g Ba(IO3)2 = mols x molar mass = estd 0.015 x about 505 = estd 7.4 g if the ppt is not soluble in water.We used 30 + 50 = 80 mL H2O.
If 4C is used we would lose 0.010 x 80/100 = 0.008 and that 7.4 would end up being 7.4-0.008 = about 7.39g before any washing.
Then if we washed with 20 mL of 4C water we would lose 0.010 x 20/100 = 0.002 and end up with 7.39 (rounded to 2 significant figures).
If we washed with 125 mL of 25C water we would lose 0.028 x 125/100 = estd 0.035g and end up with 7.39-0.0335 = estd 7.36 rounded to 2 s.f.
You should go through and recalculate using 4 s.f. to get better answers (but note that the solubilities are given only to two s.f. and not 4.). Frankly, I think this is an ill conceived problem.
Post your work if you get stuck.